Here is a "conjecture" that should be known (but I have not found any good reference to it): consider the class $\cal F$ of all $C^2([0,1])$ functions $f$ with the property $f(0)=f'(0)=f''(0)=0$ and $f(1)-1=f'(1)=f''(1)=0$. We can think of the graph of $y=f(x)$ as a transfer path between two parallel and horizontal railroads, one arriving at (0,0) and one starting at (1,1).

Such a function in $\cal F$ is the polynomial $P(x)=6x^5-15x^4+10x^3$ and the (absolute value) maximum curvature of $P$ (the general the formula for the curvature is $k_f(x)=\frac{f''(x)}{(1+f'(x)^2)^{\frac{3}{2}}}$) is a little over 4. There are other simple functions in $\cal F$ with lower maximum curvature (and it is perhaps a good exercise in Calculus III courses, to try find others, since there are infinite families in $\cal F$ that one can come up with).

For example, the piece-wise polynomial,

$$Q(x)=\begin{cases} 8x^3(1-x), \ \ \ x\in [0,\frac{1}{2}],\\ 1-8(1-x)^3x, \ \ \ \ x\in [\frac{1}{2},1],\end{cases}$$ is in $\cal F$ and it has a maximum curvature which is below $4$.

Is it true that $$m:=\underset{f\in \cal F}{inf}\left( \underset{f\in [0,1]}{max}{|k_f(x)}|\right)=2 ? $$

One can use two quarter circles of radius 1/2 and construct a path connecting the two points. The path is not quite what is required but perhaps it can be "fixed" and that will give m≤2. The other part seems to be more difficult to prove. In other words, if the curvature is less than 2, then one cannot connect those points with a curve y=f(x). If one drops this requirement and asks only for a curve in the plane, I assume the answer is m=0.

The following function, which appears from the calculation of the (normalized) area of the segment of a circle of radius $\frac{1}{2}$ in terms of its height $x$ (there is a small change of a constant in the middle though), $$f(x)=[ \arccos(1-2x)+(1-2x)(x-x^2)^{\frac{1}{2}}]/\pi,\ x\in [0,1]$$ has an inverse. Taking $g=f^{-1}$, one can check that $g\in \cal F$ and its maximum curvature which is below 2.33

  • 1
    $\begingroup$ What does lead you to think about that estimate? $\endgroup$
    – Edu
    Mar 16, 2018 at 14:59
  • $\begingroup$ I have edited the initial post to reflect this question. $\endgroup$ Mar 18, 2018 at 23:55
  • $\begingroup$ Apply the Mean Value Theorem twice $\endgroup$
    – Del
    Mar 19, 2018 at 11:03
  • $\begingroup$ I could only get that $m>2-\sqrt{2}$ with Mean Value applied several times .... $\endgroup$ Mar 26, 2018 at 19:48

1 Answer 1


We will show $m \ge 2$ and argue the actual value of $m$ is $2$.

View the graph of $f$ as a path in $\mathbb{R}^2$:

$$\gamma:\quad [0,1] \ni t\quad\mapsto\quad (x,y) = (t,f(t)) \in \mathbb{R}^2$$

Parameterize $\gamma$ by its arc-length measured from $(0,0)$, $$s(x) = \int_0^x \sqrt{1+f'(t)^2} dt$$

Let $\theta$ be the angle between tangent vector of $\gamma$ and $x$-axis, $$ (\cos\theta,\sin\theta) = \left(\frac{dx}{ds}, \frac{dy}{ds}\right) = \left(\frac{1}{\sqrt{1+f'^2}},\frac{f'}{\sqrt{1+f'^2}}\right)$$

Let $K$ be the maximum of absolute value of $\frac{d\theta}{ds}$. i.e.

$$K = K(f) \stackrel{def}{=} \max_{t\in [0,1]} \left|\frac{d\theta}{ds}\right| = \max_{t\in [0,1]} \left|\frac{f''(t)}{(1+f'(t)^2)^{3/2}}\right|$$

Let us consider what will happen when $K < 2$.

Since $f'(0) = 0 \implies \theta(0) = 0$,

$$\left|\frac{d\theta}{ds}\right| \le K \implies |\theta(s)| = \left|\int_0^s \frac{d\theta}{ds} ds\right| \le \int_0^s K ds = Ks$$

As long as $s \le \frac{\pi}{2K}$, we will have

$$\cos\theta(s) \ge \cos(Ks)\quad\text{ and }\quad |\sin\theta(s)| \le \sin(Ks)$$

Since $x(0) = y(0) = 0$, this leads to $$\begin{align} x(s) &= \int_0^s \cos\theta(s) ds \ge \int_0^s \cos(K\tau)d\tau = \frac{\sin(Ks)}{K}\tag{*1a}\\ |y(s)| &= \left|\int_0^s \sin\theta(s) ds\right| \le \int_0^s \sin(K\tau) d\tau = \frac{1-\cos(Ks)}{K}\tag{*1b} \end{align} $$

When $K < 2$, we have $\frac1K > \frac12$. $(*1a)$ tells us $x(s)$ will reach $\frac12$ at some $s = s_* < \frac{\pi}{2K}$. Together with $(*1b)$, we find $$f\left(\frac12\right) = y(s_*) \le \frac{1-\cos(Ks_*)}{K} = \frac{1 - \sqrt{1 - \sin(Ks_*)^2}}{K} \le \frac{1 - \sqrt{1 - (Kx(s_*))^2}}{K}$$ Notice $(K + \sqrt{4-K^2})^2 = 4 + 2K\sqrt{4-K^2} > 4 \implies K + \sqrt{4-K^2} > 2$, we find

$$f\left(\frac12\right) \le \frac{1 - \sqrt{1 - (Kx(s_*))^2}}{K} = \frac{2 - \sqrt{4 - K^2}}{2K} < \frac12$$

Instead of parameterize $\gamma$ using arc-length measured from $(0,0)$. we can parameterize $\gamma$ using arc-length measured from $(1,1)$. Using essentially the same argument as above but on the portion of $\gamma$ at $x \ge \frac12$, we can show that when $K < 2$, $f(\frac12) > \frac12$. From this, we can conclude it is impossible for $|\frac{d\theta}{ds}| < 2$ over the whole path. In other words $$\bbox[border:1px solid blue;padding:1em]{m = \inf_{f \in \mathcal{F}} K(f) \ge 2}$$

About the actual value of $m$, it should be $2$.

Consider following curve consisting of two quarter circular arcs:

  • one centered at $(0,\frac12)$ with radius $\frac12$ joining $(0,0)$ to $(\frac12,\frac12)$.

  • another centered at $(1,\frac12)$ with radius $\frac12$ joining $(\frac12,\frac12)$ to $(1,1)$.

This curve close to give a $f$ that we want. Aside from the point $(\frac12,\frac12)$, we have $\left|\frac{d\theta}{ds}\right| = 2$. This curve do have some minor problems. First, $f''(0), f''(1) \ne 0$. Second, $f'$ diverges at $\frac12$ and hence $f$ fails to belong to $C^2([0,1])$.

It is easy to modify $f$ to force $f''(0)$ and $f''(1)$ to vanish. If one increase the bound for $\left|\frac{d\theta}{ds}\right|$ near $x=0$ and $x=1$ a little bit, one will have enough freedom to smooth out the singularity of $f'$ at $(\frac12,\frac12)$. This suggests one can construct a $f \in C^2([0,1])$ with $K(f)$ as close to $2$ as one desire.

  • $\begingroup$ Very nice! Thanks for this clever solution. Now, it begs the question of what happens if we change $\cal F$ to $${\cal F}_t:=\{f\in C^2[0,1]| f(0)=f'(0)=0 and f(1)=1, f'(1)=t,f''(1)=0\}, \ t\in \mathbb R$$ $\endgroup$ Mar 30, 2018 at 19:13
  • 1
    $\begingroup$ @EugenIonascu given any bound $K$ of the curvature, the portion of path $\gamma$ near $(0,0)$ is avoiding two circles of radius $\frac{1}{K}$ touching $(0,0)$ (with tangent in $x$-axis). Similar thing happens for the portion near $(1,1)$, it will avoiding two circles of radius $\frac{1}{K}$ touching $(1,1)$ (with tangent in the direction $(1,t)$). For $t$ not too large, the problem should reduce to a geometry problem of finding the smallest $K$ where the two admissible region near the ends connect to each other. $\endgroup$ Mar 30, 2018 at 19:28
  • $\begingroup$ I agee! What I got is $$k_f=\frac{1}{2}\left( \cos(u)-\sin(u)+1+\sqrt{8-2(1+\sin(u))(1+\cos(u))} \right),$$ where $t=\tan u$, at least if $u\in [0,\frac{\pi}{2}]$ $\endgroup$ Mar 30, 2018 at 20:41
  • 1
    $\begingroup$ @EugenIonascu I reproduced your expression of $k_f$ and the curve looks good when implemented on GeoGebra. $\endgroup$ Mar 30, 2018 at 22:35
  • 1
    $\begingroup$ @EugenIonascu For $u \in (-\frac{\pi}{2},0)$, the curve constructed by joining the two circle arcs (with curvature given by your expression) is moving backward in $x$. Since $\gamma$ is supposed to be a graph of $f$, this cannot happen, the radius of the circles need to be smaller. The limiting condition becomes the two circles shares a common vertical tangent. $k_f$ should be $2 + \sin|u|$ when $u \in (-\frac{\pi}{2},0)$. $\endgroup$ Mar 30, 2018 at 22:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.